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Difficulty: 1 – 5

5112927

Use structure to evaluate the expression efficiently. \((-15.8) \cdot \frac{3}{7} + (-15.8) \cdot \frac{4}{7}\)

Hints

- Identify the factor shared by both terms. - Use the distributive property in reverse to factor it out. - Add the fractions before multiplying.

Solution

1. Factor out the common factor \(-15.8\): \(-15.8\left(\frac{3}{7} + \frac{4}{7}\right)\). 2. Add inside the parentheses: \(\frac{3}{7} + \frac{4}{7} = 1\). 3. Multiply: \(-15.8 \cdot 1 = -15.8\).

Answer

\(-15.8\)

5116747

Factor each expression to evaluate it efficiently. a) \(15 \cdot 48 + 15 \cdot 52\) b) \(12.5 \cdot 7 - 12.5 \cdot 3\) c) \((-20) \cdot 14 + (-20) \cdot 36\)

Hints

- Identify the factor shared by both terms in each expression. - Factor it out using the distributive property in reverse. - Evaluate the sum or difference inside the parentheses first.

Solution

1. For a), factor out \(15\): \(15(48 + 52) = 15 \cdot 100 = 1500\). 2. For b), factor out \(12.5\): \(12.5(7 - 3) = 12.5 \cdot 4 = 50\). 3. For c), factor out \(-20\): \((-20)(14 + 36) = (-20) \cdot 50 = -1000\).

Answer

a) \(1500\) b) \(50\) c) \(-1000\)

5124999

Add three consecutive integers. Use an algebraic expression to explain why their sum is always divisible by \(3\).

Hints

- How can you represent an integer and the next two integers using one variable? - Combine the terms in the sum. - How can you rewrite the result with \(3\) as a factor? - What does that factored form show about divisibility by \(3\)?

Solution

1. Let \(n\) be the first integer. The next two integers are \(n+1\) and \(n+2\). 2. Their sum is \(n+(n+1)+(n+2)=3n+3\). 3. Factoring gives \(3n+3=3(n+1)\). Because the expression is a multiple of \(3\), the sum is always divisible by \(3\).

Answer

The sum is \(n+(n+1)+(n+2)=3n+3=3(n+1)\). Therefore, it is always divisible by \(3\).

5125009

Choose any integer \(x\). Multiply it by \(5\), subtract the original number \(x\), and then add \(12\). Use an algebraic expression to show that the result is always divisible by \(4\).

Hints

- Translate each verbal step into an algebraic operation. - Combine the terms that contain \(x\). - Do the terms in the simplified expression have a common factor? - Use the distributive property to factor the expression.

Solution

1. The expression is \(5x-x+12\). 2. Combining like terms gives \(4x+12\). 3. Factoring gives \(4x+12=4(x+3)\). 4. Since the expression has \(4\) as a factor, the result is divisible by \(4\) for every integer \(x\).

Answer

The expression is \(5x-x+12=4x+12=4(x+3)\). Therefore, the result is always divisible by \(4\).

5178397

Complete each statement about how a subtraction difference changes. a) If the minuend increases by \(14\) and the subtrahend stays the same, the difference ______ by ______. b) If the minuend stays the same and the subtrahend decreases by \(9\), the difference ______ by ______. c) If both the minuend and the subtrahend increase by \(20\), then ______.

Hints

- Test each statement with a simple example such as \(20 - 10 = 10\). - What happens to the result when you subtract less? - Think about shifting both numbers the same distance on a number line.

Solution

1. Increasing the minuend by \(14\) while keeping the subtrahend fixed increases the difference by \(14\). 2. Decreasing the subtrahend by \(9\) means \(9\) less is subtracted, so the difference increases by \(9\). 3. Increasing both numbers by the same amount does not change their difference: \((m + 20) - (s + 20) = m - s\).

Answer

a) increases; \(14\) b) increases; \(9\) c) the difference stays the same

5178407

A subtraction expression is written as \(m - s = d\). Describe how \(d\) changes when both changes are made. a) Increase \(m\) by \(30\) and increase \(s\) by \(10\). b) Decrease \(m\) by \(15\) and increase \(s\) by \(15\). c) Increase \(m\) by \(12\) and decrease \(s\) by \(12\).

Hints

- Analyze the change to the minuend and the change to the subtrahend separately. - Increasing the subtrahend decreases the difference; decreasing it increases the difference. - Combine the two effects to find the net change.

Solution

1. Rewrite the expression: \((m + 30) - (s + 10) = m - s + 20\). The difference increases by \(20\). 2. Rewrite the expression: \((m - 15) - (s + 15) = m - s - 30\). The difference decreases by \(30\). 3. Rewrite the expression: \((m + 12) - (s - 12) = m - s + 24\). The difference increases by \(24\).

Answer

a) The difference increases by \(20\). b) The difference decreases by \(30\). c) The difference increases by \(24\).

5178457

Start with \(300 - 120 = 180\). Decrease the minuend by \(50\) and decrease the subtrahend by \(30\). Use how these changes affect the difference to find the new difference. Then state how much it differs from the original difference.

Hints

- Consider the effect of each change separately. - Decreasing the minuend lowers the difference, while decreasing the subtrahend raises it. - Combine the two effects, then apply the net change to the original difference.

Solution

1. Decreasing the minuend by \(50\) decreases the difference by \(50\). 2. Decreasing the subtrahend by \(30\) increases the difference by \(30\). 3. The net change is \(-50 + 30 = -20\), so the new difference is \(180 - 20 = 160\). 4. The new difference is \(20\) less than the original difference.

Answer

The new difference is \(160\). It is \(20\) less than the original difference.

5179097

A difference has a value of \(450\). The minuend is increased by \(120\), and the subtrahend is increased by \(50\). By how much does the difference change, and what is its new value?

Hints

- Analyze the two changes separately. - Increasing the subtrahend means more is subtracted. - Combine the two effects, then apply the net change to \(450\).

Solution

1. Increasing the minuend by \(120\) increases the difference by \(120\). 2. Increasing the subtrahend by \(50\) decreases the difference by \(50\). 3. The net change is \(120 - 50 = 70\), so the difference increases by \(70\). 4. The new value is \(450 + 70 = 520\).

Answer

The difference increases by \(70\), and its new value is \(520\).

5186377

Consider the expression \((560 - 140) + (320 + 180)\). Without fully evaluating the new expression, determine how its value changes if each of the four numbers is increased by \(15\). Explain your reasoning from the structure of the expression.

Hints

- Analyze each set of parentheses separately. - What happens to a difference when both numbers increase by the same amount? - What happens to a sum when both addends increase by \(15\)?

Solution

1. In \(560 - 140\), increasing both numbers by \(15\) does not change the difference. 2. In \(320 + 180\), increasing both addends by \(15\) increases the sum by \(15 + 15 = 30\). 3. Therefore, the value of the entire expression increases by \(30\). The original value is \(920\), and the new value is \(950\).

Answer

The value increases by \(30\).

5186387

Consider the expression \((1200 + 800) - (600 - 200)\). Determine how its value changes if each of the four numbers is increased by \(25\). Explain your reasoning from the structure of the expression.

Hints

- Determine how the first set of parentheses changes. - Determine how the second set of parentheses changes. - Use those two changes to determine the change in the entire expression.

Solution

1. Increasing both addends in \(1200 + 800\) by \(25\) increases that sum by \(50\). 2. Increasing both numbers in \(600 - 200\) by \(25\) leaves that difference unchanged. 3. Therefore, the value of the entire expression increases by \(50\). The original value is \(1600\), and the new value is \(1650\).

Answer

The value increases by \(50\).

5192107

Compare expressions \(A\) and \(B\). Which has the greater value? Explain without carrying out the full calculations. \(A = (333 \cdot 77) \div 77\) \(B = (330 + 77) - 77\)

Hints

- Identify the inverse operations in each expression. - Determine which operations undo each other. - Compare the starting values after simplifying the structure.

Solution

1. In expression \(A\), multiplying by \(77\) and then dividing by \(77\) undo each other, so \(A = 333\). 2. In expression \(B\), adding \(77\) and then subtracting \(77\) undo each other, so \(B = 330\). 3. Since \(333 > 330\), expression \(A\) has the greater value.

Answer

Expression \(A\) is greater because \(A = 333\) and \(B = 330\).

5193267

Rewrite each expression to make it easy to evaluate mentally. a) \(25 \cdot 17 \cdot 4\) b) \(125 \cdot 11 \cdot 8\) c) \(5 \cdot 42 \cdot 20\) d) \(63 \cdot 8 + 63 \cdot 2\)

Hints

- Regroup factors to make \(10\), \(100\), or \(1000\). - In part d, identify the common factor. - Use the distributive property in reverse when a factor appears in both products.

Solution

1. For a), regroup: \((25 \cdot 4) \cdot 17 = 100 \cdot 17 = 1700\). 2. For b), regroup: \((125 \cdot 8) \cdot 11 = 1000 \cdot 11 = 11{,}000\). 3. For c), regroup: \((5 \cdot 20) \cdot 42 = 100 \cdot 42 = 4200\). 4. For d), factor out the common factor: \(63(8 + 2) = 63 \cdot 10 = 630\).

Answer

a) \(1700\) b) \(11{,}000\) c) \(4200\) d) \(630\)

5196427

Write and evaluate an expression for this instruction: Add the product of \(47\) and \(19\) to the product of \(47\) and \(81\). Factor out the common factor to calculate efficiently.

Hints

- Translate each phrase “the product of” into multiplication. - Identify the factor that appears in both products. - Factor it out before adding the remaining factors.

Solution

1. The expression is \(47 \cdot 19 + 47 \cdot 81\). 2. Factor out the common factor \(47\): \(47(19 + 81)\). 3. Evaluate: \(47 \cdot 100 = 4700\).

Answer

\(47 \cdot 19 + 47 \cdot 81 = 47(19 + 81) = 4700\)

5199477

Name the property or properties used in each rewrite. a) \(28 + 57 + 72 = 28 + 72 + 57\) b) \((4 \cdot 13) \cdot 25 = 13 \cdot (4 \cdot 25)\) c) \(9 \cdot 14 + 9 \cdot 6 = 9(14 + 6)\)

Hints

- Check whether the order of terms changes. - Check whether the grouping changes. - Check whether a common factor is distributed or factored out.

Solution

1. In a), the order of \(57\) and \(72\) changes, so the commutative property of addition is used. 2. In b), the factors are reordered and regrouped, so the commutative and associative properties of multiplication are used. 3. In c), the common factor \(9\) is factored out, so the distributive property is used in reverse.

Answer

a) Commutative property of addition b) Commutative and associative properties of multiplication c) Distributive property

5225339

Use algebraic expressions to describe and analyze each set of numbers. 1. Write an expression for all whole numbers that leave a remainder of \(1\) when divided by \(4\). Let \(n=0,1,2,\ldots\). 2. Explain why \(2(k+5)\) is even for every integer \(k\). 3. A student claims that every number of the form \(3n+6\) is divisible by \(3\). Verify the claim by factoring the expression.

Hints

- A number with remainder \(1\) is one more than a multiple of the divisor. - Use the definition of an even number. - Factoring can reveal whether every value of an expression has a particular factor.

Solution

1. Multiples of \(4\) have the form \(4n\). Adding \(1\) gives all numbers with remainder \(1\), so the expression is \(4n+1\). 2. Because \(k\) is an integer, \(k+5\) is also an integer. Therefore, \(2(k+5)\) is \(2\) times an integer and must be even. 3. Factor out \(3\): \(3n+6=3(n+2)\). The expression is a multiple of \(3\) for every integer \(n\), so the claim is true.

Answer

1. \(4n+1\) 2. It is \(2\) times an integer, so it is even. 3. The claim is true because \(3n+6=3(n+2)\).

5230037

Consider three numbers spaced \(1\) unit apart. If the middle number is \(n\), the three numbers are \(n-1\), \(n\), and \(n+1\). 1. Write and simplify an expression for their sum. 2. What relationship does the simplified expression describe? 3. Verify the relationship for \(n=15\) and \(n=2.5\) by comparing each direct sum with the value of your expression.

Hints

- What happens when you remove the parentheses from the sum? - What is the sum of \(-1\) and \(1\)? - How many copies of \(n\) remain? - Substitute each given value into both the three-number sum and the simplified expression.

Solution

1. The sum is \((n-1)+n+(n+1)\). Combining like terms gives \(3n\). 2. The expression shows that the sum of three numbers spaced \(1\) unit apart is three times the middle number. 3. For \(n=15\), the numbers are \(14\), \(15\), and \(16\). Their sum is \(14+15+16=45\), and \(3\cdot15=45\). 4. For \(n=2.5\), the numbers are \(1.5\), \(2.5\), and \(3.5\). Their sum is \(1.5+2.5+3.5=7.5\), and \(3\cdot2.5=7.5\).

Answer

1. \(3n\) 2. The sum is three times the middle number. 3. Both methods give \(45\) when \(n=15\) and \(7.5\) when \(n=2.5\).

5230057

Five consecutive positive integers have middle number \(m\), where \(m\ge3\). a) Write an expression for each of the five integers in terms of \(m\). b) Find their sum and simplify the expression completely.

Hints

- How do you represent the integers immediately before and after \(m\)? - Try a numerical example with middle number \(10\). - What is the sum of \(-2\), \(-1\), \(1\), and \(2\)?

Solution

1. The two integers before \(m\) are \(m-2\) and \(m-1\), and the two integers after it are \(m+1\) and \(m+2\). 2. The sum is \((m-2)+(m-1)+m+(m+1)+(m+2)\). 3. Combining like terms gives \(5m\), because the constants sum to \(0\).

Answer

a) \(m-2\), \(m-1\), \(m\), \(m+1\), and \(m+2\) b) \(5m\)

5236019

Use variables to prove that the sum of an even integer and an odd integer is always odd.

Hints

- Represent an even integer as \(2n\). - Represent an odd integer as \(2m+1\). - Rewrite the sum so that a factor of \(2\) is visible. - An odd integer has remainder \(1\) when divided by \(2\).

Solution

1. Write an even integer as \(2n\) and an odd integer as \(2m+1\), where \(m\) and \(n\) are integers. 2. Add the expressions: \(2n+(2m+1)=2n+2m+1\). 3. Factor out \(2\): \(2n+2m+1=2(n+m)+1\). 4. Since \(n+m\) is an integer, the sum has the form \(2k+1\), so it is odd.

Answer

\(2n+(2m+1)=2(n+m)+1\). This has the form of an odd integer, so the sum is always odd.

5317467

Luisa is building \(4\) identical wooden picture frames. Each frame needs \(2\) long strips and \(2\) short strips. The two calculation trees show the strip lengths, in meters, and two different methods. a) Explain what each step in tree a) and tree b) represents in the situation. b) Find every missing value in both trees and determine the total length of wood needed. c) Name the property that guarantees both methods give the same result. Write the expression represented by each tree.

Figure for problem 531746

Hints

- Determine how many long and short strips are needed for all four frames. - Work from the leaves toward the root of each tree. - One tree calculates one frame first; the other calculates each strip type for all frames. - Which property distributes multiplication across a sum?

Solution

1. Tree a) finds the wood needed for one frame: \(2\cdot1.2=2.4\) meters of long strips and \(2\cdot0.85=1.7\) meters of short strips. It adds these to get \(4.1\) meters per frame, then multiplies by \(4\). 2. Tree a) gives \(4(2.4+1.7)=4\cdot4.1=16.4\,\text{m}\). 3. Tree b) uses \(8\) long strips and \(8\) short strips for all four frames. The partial lengths are \(8\cdot1.2=9.6\,\text{m}\) and \(8\cdot0.85=6.8\,\text{m}\). 4. Tree b) gives \(9.6+6.8=16.4\,\text{m}\). 5. The distributive property shows the methods are equivalent. The expressions are \(4(2\cdot1.2+2\cdot0.85)\) and \(8\cdot1.2+8\cdot0.85\).

Answer

a) Tree a) finds the wood for one frame and then multiplies by \(4\). Tree b) finds the total lengths of all long and all short strips separately, then adds them. b) Tree a) has missing values \(2.4\), \(1.7\), \(4.1\), and \(16.4\). Tree b) has missing values \(9.6\), \(6.8\), and \(16.4\). The total length is \(16.4\,\text{m}\). c) Distributive property; \(4(2\cdot1.2+2\cdot0.85)\) and \(8\cdot1.2+8\cdot0.85\).

5546937

A concert ticket originally costs \(p\) dollars. A service charge equal to \(20\%\) of the ticket price is added. Rewrite \(p+0.20p\) as a single product. Explain what the multiplier reveals about the new price.

Hints

- Write \(p\) as \(1p\). - Combine the coefficients of the same variable. - Interpret a multiplier greater than \(1\) as the original amount plus an increase.

Solution

1. Combine the coefficients of \(p\): \(p+0.20p=1.20p\). 2. The multiplier \(1.20\) means the new price is \(120\%\) of the original price, which is a \(20\%\) increase.

Answer

\(1.20p\). The multiplier \(1.20\) reveals that the new price is \(120\%\) of the original, or \(20\%\) more.

5546947

After a sale discount, a jacket costs \(0.80p\) dollars, where \(p\) is its original price. Rewrite \(0.80p\) in the form \(p-rp\). Find \(r\) and explain what percent discount the expression reveals.

Hints

- Think of \(p\) as \(1p\). - Find the amount that must be subtracted from \(1\) to get \(0.80\). - Convert the decimal coefficient you find into a percent.

Solution

1. Since \(0.80=1-0.20\), \(0.80p=p-0.20p\). 2. Thus \(r=0.20\). 3. The term \(0.20p\) is \(20\%\) of the original price, so the expression reveals a \(20\%\) discount.

Answer

\(0.80p=p-0.20p\), so \(r=0.20\). The discount is \(20\%\).

5101257

A city has a population of \(36{,}000\). During the first year, the population grows by \(0.5\%\). a) What is the population after the first year? b) During the second year, the new population grows by \(p\%\). Write an expression for the population after the second year.

Hints

- Treat the two yearly changes separately. - The second-year increase is based on the population after the first year. - Write the percent increase using the variable \(p\). - Leave the variable in the final expression.

Solution

1. For part a, the first-year increase is \(36{,}000 \cdot 0.005 = 180\). The population after the first year is \(36{,}000 + 180 = 36{,}180\). 2. For part b, the second-year increase is \(36{,}180 \cdot \frac{p}{100} = 361.8p\). 3. Therefore, the population after the second year is \(36{,}180 + 361.8p\), or equivalently \(36{,}180\left(1 + \frac{p}{100}\right)\).

Answer

a) The population after the first year is \(36{,}180\). b) The population after the second year is \(36{,}180\left(1 + \frac{p}{100}\right)\).

5108807

Use number structure, known squares, or helpful decompositions to solve each problem mentally. a) \(196\div14\) b) \(18\cdot22\) c) \(1001\div7\) d) \(-13\cdot13\)

Hints

- Look for familiar square numbers. - Can you rewrite two factors as numbers equally far from the same multiple of ten? - Break a large dividend into multiples of the divisor that are easy to divide mentally. - Decide the sign before using a known square.

Solution

1. For a), recognize that \(14^2=196\), so \(196\div14=14\). 2. For b), use \((20-2)(20+2)=20^2-2^2=400-4=396\). 3. For c), decompose \(1001=700+280+21\). Then \(1001\div7=100+40+3=143\). 4. For d), the product is negative and \(13^2=169\), so \(-13\cdot13=-169\).

Answer

a) \(14\) b) \(396\) c) \(143\) d) \(-169\)

5113187

Factor each expression to reveal an efficient calculation. Show your work. a) \(\frac{4}{7} \cdot (-13.2) + \frac{4}{7} \cdot (-0.8)\) b) \(-15.5 \cdot \frac{2}{3} + 6.5 \cdot \frac{2}{3}\)

Hints

- Find the factor shared by both terms in each expression. - Factor it out using the distributive property in reverse. - Combine the numbers inside the parentheses before multiplying.

Solution

1. For a), factor out \(\frac{4}{7}\): \(\frac{4}{7}(-13.2 - 0.8) = \frac{4}{7}(-14) = -8\). 2. For b), factor out \(\frac{2}{3}\): \(\frac{2}{3}(-15.5 + 6.5) = \frac{2}{3}(-9) = -6\).

Answer

a) \(-8\) b) \(-6\)

5113277

Analyze how parentheses change both the structure and value of an expression. Describe the overall structure of each expression and evaluate it. a) \((-0.4)^2-0.16\div0.4\) b) \(\left((-0.4)^2-0.16\right)\div0.4\)

Hints

- Apply powers before multiplication or division, and multiplication or division before addition or subtraction. - How do the parentheses in b) change which operation is performed last? - Pay attention to the sign when squaring a negative number.

Solution

1. For a), the last operation is subtraction, so the expression is a difference. Compute \((-0.4)^2=0.16\) and \(0.16\div0.4=0.4\). Then \(0.16-0.4=-0.24\). 2. For b), the last operation is division, so the expression is a quotient. Inside the parentheses, \((-0.4)^2-0.16=0.16-0.16=0\). Then \(0\div0.4=0\).

Answer

a) Structure: difference; value: \(-0.24\) b) Structure: quotient; value: \(0\)

5113427

Consider the expression \(T=\frac{-3.6+1.2}{-3.6\div1.2}\). a) Describe the complete structure of the expression using terms such as sum and quotient. b) Evaluate the expression.

Hints

- A fraction bar acts like division and groups the numerator and denominator. - Which operation is performed last in the entire expression? - Evaluate the numerator and denominator separately. - Use sign rules when dividing rational numbers.

Solution

1. The overall expression is a quotient. Its numerator is the sum of \(-3.6\) and \(1.2\), and its denominator is the quotient of \(-3.6\) and \(1.2\). 2. Numerator: \(-3.6+1.2=-2.4\). 3. Denominator: \(-3.6\div1.2=-3\). 4. Final division: \(-2.4\div(-3)=0.8\).

Answer

a) The expression is the quotient of the sum \(-3.6+1.2\) and the quotient \(-3.6\div1.2\). b) \(T=0.8\)

5116767

Evaluate each expression by factoring to reveal a simpler calculation. a) \(720 \div 9 - 180 \div 9\) b) \(1.2 \cdot 15 + 1.2 \cdot 5 - 1.2 \cdot 10\)

Hints

- In a), both terms are divided by the same number. - In b), all three terms share the same factor. - Factor first, then evaluate the expression inside the parentheses.

Solution

1. For a), view division by \(9\) as multiplication by \(\frac{1}{9}\) and factor: \((720 - 180) \div 9 = 540 \div 9 = 60\). 2. For b), factor out \(1.2\): \(1.2(15 + 5 - 10) = 1.2 \cdot 10 = 12\).

Answer

a) \(60\) b) \(12\)

5117327

Consider expressions \(A\) and \(B\): \(A=8.4\div(1.2+0.9)\) \(B=8.4\div1.2+0.9\) Evaluate both expressions and briefly explain why their values are different.

Hints

- What order of operations applies when there are no parentheses? - How do parentheses change which quantity is the divisor? - Evaluate each expression step by step.

Solution

1. For \(A\), evaluate the parentheses first: \(1.2+0.9=2.1\). Then \(8.4\div2.1=4\). 2. For \(B\), division comes before addition: \(8.4\div1.2=7\), then \(7+0.9=7.9\). 3. The parentheses in \(A\) make the entire sum the divisor, while in \(B\) only \(1.2\) is the divisor.

Answer

\(A=4\); \(B=7.9\). The parentheses change the structure and order of operations.

5122607

Factor each expression using the distributive property, then evaluate it. a) \(27 \cdot 13 + 27 \cdot 87\) b) \(0.4 \cdot 35 - 0.4 \cdot 15\) c) \(\frac{3}{8} \cdot 17 - \frac{3}{8} \cdot 9\) d) \((-12) \cdot 0.25 + (-8) \cdot 0.25\)

Hints

- Identify the factor shared by every term in each expression. - Factor first, then evaluate the sum or difference inside the parentheses. - Keep negative signs attached to their factors.

Solution

1. For a), factor out \(27\): \(27(13 + 87) = 27 \cdot 100 = 2700\). 2. For b), factor out \(0.4\): \(0.4(35 - 15) = 0.4 \cdot 20 = 8\). 3. For c), factor out \(\frac{3}{8}\): \(\frac{3}{8}(17 - 9) = \frac{3}{8} \cdot 8 = 3\). 4. For d), factor out \(0.25\): \(0.25(-12 - 8) = 0.25 \cdot (-20) = -5\).

Answer

a) \(2700\) b) \(8\) c) \(3\) d) \(-5\)

5122737

A day has \(24\) hours. A student plans \(8.5\) hours for sleep, \(6.5\) hours for school, \(1.5\) hours for homework, and \(t_H\) hours for hobbies. Let \(F\) represent the remaining free time. a) Write and simplify an expression for \(F\) in terms of \(t_H\). b) School time increases to \(7.5\) hours, while sleep decreases to \(8.0\) hours. How does this change \(F\)? Justify your answer by comparing the expressions.

Hints

- Add all fixed time amounts first. - Increasing the total time used for other activities decreases the free time. - Write a new expression with the changed values, then compare it with the original expression.

Solution

1. The original free time is \(F=24-(8.5+6.5+1.5+t_H)\). 2. Combine the fixed times: \(8.5+6.5+1.5=16.5\). Thus, \(F=24-16.5-t_H=7.5-t_H\). 3. With the changes, \(F_{\text{new}}=24-(8.0+7.5+1.5+t_H)\). 4. Since \(8.0+7.5+1.5=17.0\), \(F_{\text{new}}=7.0-t_H\). This is \(0.5\) hour less than the original expression.

Answer

a) \(F=7.5-t_H\) b) The free time decreases by \(0.5\) hour because the fixed-time total increases from \(16.5\) hours to \(17.0\) hours.

5124697

Place exactly one pair of parentheses in \(40-20-10-5\). The parentheses must contain at least two numbers. a) Find the value without parentheses. b) Place the parentheses so the value does not change. c) Place the parentheses so the value is \(25\).

Hints

- Without parentheses, perform subtraction from left to right. - A minus sign before parentheses changes how the enclosed difference affects the result. - Test possible groups of consecutive numbers.

Solution

1. Without parentheses, calculate from left to right: \(40-20-10-5=5\). 2. One placement that keeps the value is \((40-20)-10-5=5\). Another is \((40-20-10)-5=5\). 3. To obtain \(25\), use \(40-(20-10)-5=25\).

Answer

a) \(5\) b) \((40-20)-10-5\) or \((40-20-10)-5\) c) \(40-(20-10)-5\)

5124707

Consider the expression \(50-25-10-5\). Determine whether placing exactly one pair of parentheses can make the value \(40\). The parentheses must enclose at least two numbers. Justify your answer by evaluating every possible placement.

Hints

- Count all the ways one pair of parentheses can enclose consecutive numbers in a four-number expression. - Evaluate each possible placement carefully. - Remember to simplify the expression inside parentheses first.

Solution

1. Parentheses around the first two numbers give \((50-25)-10-5=10\). 2. Parentheses around the first three numbers give \((50-25-10)-5=10\). 3. Parentheses around the entire expression give \((50-25-10-5)=10\). 4. Parentheses around the second and third numbers give \(50-(25-10)-5=30\). 5. Parentheses around the second through fourth numbers give \(50-(25-10-5)=40\). 6. Parentheses around the third and fourth numbers give \(50-25-(10-5)=20\). 7. Therefore, it is possible. The required placement is \(50-(25-10-5)\).

Answer

Yes. The placement \(50-(25-10-5)\) gives \(40\).

5124969

Lukas claims, “When I add four consecutive whole numbers, the sum is always divisible by \(4\).” a) Test Lukas’s claim with two examples. b) Write an expression for the sum of four consecutive whole numbers. Let \(n\) be the smallest number. c) Use the expression to explain why Lukas’s claim is false.

Hints

- How can you represent all four numbers using one variable? - What do you get when you combine like terms? - How can rewriting the expression show its remainder when divided by \(4\)? - How many counterexamples are needed to disprove a universal claim?

Solution

1. Two examples are \(1+2+3+4=10\) and \(5+6+7+8=26\). Neither sum is divisible by \(4\). 2. The four numbers are \(n\), \(n+1\), \(n+2\), and \(n+3\), so their sum is \(n+(n+1)+(n+2)+(n+3)\). 3. Combining like terms gives \(4n+6\). 4. Rewrite the expression as \(4(n+1)+2\). It always leaves a remainder of \(2\) when divided by \(4\), so the sum is never divisible by \(4\).

Answer

a) For example, \(1+2+3+4=10\) and \(5+6+7+8=26\). Both examples disprove the claim. b) \(4n+6\) c) Since \(4n+6=4(n+1)+2\), the sum always has a remainder of \(2\) when divided by \(4\).

5124979

Consider three consecutive even integers, such as \(2\), \(4\), and \(6\), or \(20\), \(22\), and \(24\). Use an algebraic expression to show that the sum of any three consecutive even integers is divisible by \(6\).

Hints

- How can you represent any even integer using a variable? - How do you get from one even integer to the next? - Try to rewrite the sum with \(6\) as a factor.

Solution

1. Let the first even integer be \(2n\), where \(n\) is an integer. 2. The next two consecutive even integers are \(2n+2\) and \(2n+4\). 3. Their sum is \(2n+(2n+2)+(2n+4)=6n+6\). 4. Factoring gives \(6n+6=6(n+1)\). Because the sum is a multiple of \(6\), it is divisible by \(6\).

Answer

The integers can be written as \(2n\), \(2n+2\), and \(2n+4\). Their sum is \(6n+6=6(n+1)\), so it is always divisible by \(6\).

5124987

A hundreds chart lists the numbers from \(1\) through \(100\) in rows of ten. <table><tbody> <tr><td>\(1\)</td><td>\(2\)</td><td>\(3\)</td><td>\(4\)</td><td>\(5\)</td><td>\(6\)</td><td>\(7\)</td><td>\(8\)</td><td>\(9\)</td><td>\(10\)</td></tr> <tr><td>\(11\)</td><td>\(12\)</td><td>\(13\)</td><td>\(14\)</td><td>\(15\)</td><td>\(16\)</td><td>\(17\)</td><td>\(18\)</td><td>\(19\)</td><td>\(20\)</td></tr> <tr><td>\(21\)</td><td>\(22\)</td><td>\(23\)</td><td>\(24\)</td><td>\(25\)</td><td>\(26\)</td><td>\(27\)</td><td>\(28\)</td><td>\(29\)</td><td>\(30\)</td></tr> <tr><td>\(31\)</td><td>\(32\)</td><td>\(33\)</td><td>\(34\)</td><td>\(35\)</td><td>\(36\)</td><td>\(37\)</td><td>\(38\)</td><td>\(39\)</td><td>\(40\)</td></tr> <tr><td>\(41\)</td><td>\(42\)</td><td>\(43\)</td><td>\(44\)</td><td>\(45\)</td><td>\(46\)</td><td>\(47\)</td><td>\(48\)</td><td>\(49\)</td><td>\(50\)</td></tr> <tr><td>\(51\)</td><td>\(52\)</td><td>\(53\)</td><td>\(54\)</td><td>\(55\)</td><td>\(56\)</td><td>\(57\)</td><td>\(58\)</td><td>\(59\)</td><td>\(60\)</td></tr> <tr><td>\(61\)</td><td>\(62\)</td><td>\(63\)</td><td>\(64\)</td><td>\(65\)</td><td>\(66\)</td><td>\(67\)</td><td>\(68\)</td><td>\(69\)</td><td>\(70\)</td></tr> <tr><td>\(71\)</td><td>\(72\)</td><td>\(73\)</td><td>\(74\)</td><td>\(75\)</td><td>\(76\)</td><td>\(77\)</td><td>\(78\)</td><td>\(79\)</td><td>\(80\)</td></tr> <tr><td>\(81\)</td><td>\(82\)</td><td>\(83\)</td><td>\(84\)</td><td>\(85\)</td><td>\(86\)</td><td>\(87\)</td><td>\(88\)</td><td>\(89\)</td><td>\(90\)</td></tr> <tr><td>\(91\)</td><td>\(92\)</td><td>\(93\)</td><td>\(94\)</td><td>\(95\)</td><td>\(96\)</td><td>\(97\)</td><td>\(98\)</td><td>\(99\)</td><td>\(100\)</td></tr> </tbody></table> Choose a number \(x\) that is not in the first or last row or column. Mark \(x\) and the four spaces directly above, below, left, and right of it. Use an expression to show that the sum of the five numbers in this cross is always \(5\) times the center number \(x\).

Hints

- How does a number change when you move one space left or right on a hundreds chart? - How does it change when you move one row up or down? - Write an expression for all five spaces, then combine like terms.

Solution

1. The numbers to the left and right of \(x\) are \(x-1\) and \(x+1\). The numbers above and below \(x\) are \(x-10\) and \(x+10\). 2. The sum is \(x+(x-1)+(x+1)+(x-10)+(x+10)\). 3. The constants cancel in opposite pairs, so the expression simplifies to \(5x\). 4. Therefore, the sum is always \(5\) times the center number.

Answer

The five numbers are \(x-10\), \(x-1\), \(x\), \(x+1\), and \(x+10\). Their sum simplifies to \(5x\), so it is always \(5\) times the center number.

5126237

A mental math performer at a carnival says, “Choose a secret number, double it, add \(10\), and divide the result by \(2\). Tell me your final result.” a) Let the starting number be \(x\). Write and simplify an expression for the final result. b) One participant says the final result is \(42\). What was the starting number? c) Describe the quick mental rule the performer can use to recover the starting number.

Hints

- Write an expression that follows the operations in order. - Can you divide each term in the numerator by \(2\)? - Once you know how the final result compares with the starting number, reverse that change.

Solution

1. The final result is \(\frac{2x+10}{2}\). 2. Dividing both terms in the numerator by \(2\) gives \(x+5\). 3. For a final result of \(42\), solve \(x+5=42\). Subtracting \(5\) gives \(x=37\). 4. Since the final result is always \(5\) more than the starting number, the performer subtracts \(5\) from the reported result.

Answer

a) \(\frac{2x+10}{2}=x+5\) b) \(37\) c) Subtract \(5\) from the final result.

5128197

Compare expressions \(A\) and \(B\). Which has the greater value? Show your work. \(A=-3\cdot(2.4-5.4)\) \(B=-3\cdot2.4-5.4\)

Hints

- How do the parentheses change the order of operations? - Compare the structures of the two expressions carefully. - Predict the sign of each result before calculating.

Solution

1. For \(A\), evaluate the parentheses first: \(2.4-5.4=-3\). Then \(-3\cdot(-3)=9\). 2. For \(B\), multiply first: \(-3\cdot2.4=-7.2\). Then \(-7.2-5.4=-12.6\). 3. Since \(9>-12.6\), expression \(A\) has the greater value.

Answer

Expression \(A\) has the greater value because \(9>-12.6\).

5133887

Consider the expressions, where \(b\neq0\), \(c\neq0\), and \(d\neq0\): \(T_1=\frac{a}{b}\div\frac{c}{d}\) \(T_2=\frac{a\cdot d}{b\cdot c}\) a) Evaluate \(T_1\) and \(T_2\) for \(a=2\), \(b=3\), \(c=4\), and \(d=5\). b) Rewrite \(T_1\) to explain why the two expressions are equivalent for all allowed values of the variables.

Hints

- Replace division by a fraction with multiplication by its reciprocal. - Evaluate each expression separately for the given values. - For part b), rewrite \(T_1\) symbolically and compare its numerator and denominator with \(T_2\).

Solution

1. For a), \(T_1=\frac{2}{3}\div\frac{4}{5}=\frac{2}{3}\cdot\frac{5}{4}=\frac{5}{6}\). 2. Also, \(T_2=\frac{2\cdot5}{3\cdot4}=\frac{5}{6}\). 3. For b), dividing by \(\frac{c}{d}\) is equivalent to multiplying by its reciprocal \(\frac{d}{c}\): \(\frac{a}{b}\div\frac{c}{d}=\frac{a}{b}\cdot\frac{d}{c}=\frac{ad}{bc}\). 4. This is exactly \(T_2\), so the expressions are equivalent whenever the stated nonzero conditions hold.

Answer

a) \(T_1=T_2=\frac{5}{6}\) b) \(\frac{a}{b}\div\frac{c}{d}=\frac{a}{b}\cdot\frac{d}{c}=\frac{ad}{bc}\), so the expressions are equivalent.

5178467

In a subtraction expression, the minuend is increased by \(45\). How must the subtrahend change so that the difference is \(60\) greater than it was originally?

Hints

- Find the effect of increasing the minuend by \(45\). - Determine how much additional change is needed to reach \(+60\). - Decide whether increasing or decreasing the subtrahend produces that change.

Solution

1. Increasing the minuend by \(45\) increases the difference by \(45\). 2. The desired total increase is \(60\), so an additional increase of \(60 - 45 = 15\) is needed. 3. Decreasing the subtrahend by \(15\) increases the difference by \(15\): \((m + 45) - (s - 15) = m - s + 60\).

Answer

The subtrahend must decrease by \(15\).

5179107

In a subtraction expression, the subtrahend is \(2458\). It is \(1035\) less than the minuend. a) Find the minuend. b) The minuend stays fixed. By how much must the subtrahend decrease for the difference to increase by \(150\)?

Hints

- Add the difference between the numbers to the smaller number to find the minuend. - When the minuend stays fixed, subtracting less makes the difference greater. - In part b), decide whether you need the original value of the difference.

Solution

1. Since the subtrahend is \(1035\) less than the minuend, the minuend is \(2458 + 1035 = 3493\). 2. With the minuend fixed, decreasing the subtrahend by \(150\) increases the difference by \(150\).

Answer

a) The minuend is \(3493\). b) The subtrahend must decrease by \(150\).

5179817

Three students evaluate \(1000 - 250 - 150 - 100\). - **Leo:** \(1000 - 250 = 750\), \(750 - 150 = 600\), \(600 - 100 = 500\). - **Mia:** \(250 + 150 + 100 = 500\), then \(1000 - 500 = 500\). - **Noah:** \(1000 - 250 = 750\), \(150 - 100 = 50\), then \(750 - 50 = 700\). Decide which methods are correct. Explain Noah’s error if his method is incorrect.

Hints

- Subtraction of several positive amounts can be rewritten by adding the amounts and subtracting their total. - Check whether each rewrite is equivalent to the original expression. - Compare \(-150 - 100\) with \(-(150 - 100)\).

Solution

1. Leo evaluates the subtraction from left to right and obtains \(500\), so his method is correct. 2. Mia combines all three amounts being subtracted: \(1000 - 250 - 150 - 100 = 1000 - (250 + 150 + 100) = 500\). Her method is also correct. 3. Noah replaces subtracting both \(150\) and \(100\) with subtracting their difference. However, \(-150 - 100\) is not equal to \(-(150 - 100)\). His result is incorrect.

Answer

Leo and Mia are correct. Noah is incorrect because he subtracts \(150 - 100\) instead of subtracting both \(150\) and \(100\).

5186327

Find the sum \(1 + 2 + 3 + \dots + 79 + 80\) by pairing terms. 1) Name the two properties that let you reorder and regroup the addends. 2) How many pairs are formed when you pair the first term with the last term, the second term with the next-to-last term, and so on? 3) What is the sum of each pair? 4) Find the total sum.

Hints

- Compare the sums of the first and last terms and of the second and next-to-last terms. - Divide the number of terms by \(2\) to find the number of pairs. - Multiply the number of pairs by the sum of each pair.

Solution

1. The commutative property allows the terms to be reordered, and the associative property allows them to be regrouped. 2. There are \(80 \div 2 = 40\) pairs. 3. Each pair has a sum of \(81\), since \(1 + 80 = 81\), \(2 + 79 = 81\), and so on. 4. Therefore, the total is \(40 \cdot 81 = 3240\).

Answer

1) Commutative property and associative property 2) \(40\) pairs 3) \(81\) 4) \(3240\)

5186337

An apple seller stacks apples in a triangular display. The top row has \(1\) apple, and each row below it has one more apple than the row above. The bottom row has \(40\) apples. 1) Write an expression for the total number of apples. 2) Find the total efficiently by pairing terms. 3) Explain why \(20 \cdot 41\) gives the total.

Hints

- Write the first few and last few terms of the sum. - Pair terms from opposite ends of the list. - Determine how many pairs are formed and the sum of each pair.

Solution

1. The total is represented by \(1 + 2 + 3 + \dots + 40\). 2. Pair the first and last terms: \(1 + 40 = 41\), \(2 + 39 = 41\), and so on. Since \(40 \div 2 = 20\), there are \(20\) pairs. 3. Each pair has a sum of \(41\), so the total is \(20 \cdot 41 = 820\).

Answer

1) \(1 + 2 + 3 + \dots + 40\) 2) \(820\) apples 3) There are \(20\) pairs, and each pair has a sum of \(41\).

5186347

Find the sum of all even numbers from \(2\) through \(60\): \(2 + 4 + 6 + \dots + 58 + 60\) 1) How many terms are in the sum? 2) Pair terms that have the same sum. How many pairs are there, and what is the sum of each pair? 3) Find the total.

Hints

- Divide \(60\) by \(2\) to count the even numbers. - Pair the smallest term with the largest term. - Multiply the number of pairs by the sum of each pair.

Solution

1. There are \(60 \div 2 = 30\) even numbers from \(2\) through \(60\). 2. Pair terms from opposite ends: \(2 + 60 = 62\), \(4 + 58 = 62\), and so on. The \(30\) terms form \(30 \div 2 = 15\) pairs. 3. The total is \(15 \cdot 62 = 930\).

Answer

1) \(30\) terms 2) \(15\) pairs; each pair has a sum of \(62\) 3) \(930\)

5187147

Rewrite the expression to make it easy to evaluate. Keep each operation sign with its number when you reorder the terms. \(864 + 257 - 364 + 43\)

Hints

- Look for two numbers with matching final digits that make an easy difference. - Look for two addends that make a multiple of \(100\). - When reordering, move each operation sign with the term that follows it.

Solution

1. Reorder the terms: \(864 - 364 + 257 + 43\). 2. Group convenient pairs: \((864 - 364) + (257 + 43)\). 3. Evaluate: \(500 + 300 = 800\).

Answer

\(800\)

5187157

Use the distributive property to evaluate the expression efficiently. Show how you factor out the common factor. \(14 \cdot 63 + 14 \cdot 37\)

Hints

- Identify the factor that appears in both products. - Factor that number out of the sum. - Add the remaining factors before multiplying.

Solution

1. Both products have a common factor of \(14\). 2. Factor out \(14\): \(14 \cdot (63 + 37)\). 3. Evaluate: \(14 \cdot 100 = 1400\).

Answer

\(14 \cdot 63 + 14 \cdot 37 = 14 \cdot (63 + 37) = 1400\)

5191307

Rewrite the repeated addition as a sum of two products. Then use a common factor to evaluate efficiently. \(18 + 18 + 18 + 18 + 18 + 12 + 12 + 12 + 12 + 12\)

Hints

- Count how many times each addend appears. - Write each repeated addend as a product. - Look for a common factor in the two products.

Solution

1. The number \(18\) appears five times, and \(12\) appears five times, so the expression is \(5 \cdot 18 + 5 \cdot 12\). 2. Factor out the common factor \(5\): \(5(18 + 12)\). 3. Evaluate: \(5 \cdot 30 = 150\).

Answer

\(5 \cdot 18 + 5 \cdot 12 = 5(18 + 12) = 150\)

5191347

Rewrite each expression in a form that is easy to evaluate mentally. a) \(4 \cdot 19 \cdot 25\) b) \(80 \cdot 600\) c) \(54{,}000 \div 90\) d) \(2 \cdot 37 \cdot 50\)

Hints

- Look for factors that make \(10\), \(100\), or \(1000\). - Use place value to rewrite products that contain trailing zeros. - In the quotient, divide both numbers by the same power of \(10\).

Solution

1. For a), regroup the factors: \((4 \cdot 25) \cdot 19 = 100 \cdot 19 = 1900\). 2. For b), write the factors by place value: \(80 \cdot 600 = 8 \cdot 6 \cdot 1000 = 48{,}000\). 3. For c), divide the dividend and divisor by \(10\): \(54{,}000 \div 90 = 5400 \div 9 = 600\). 4. For d), regroup the factors: \((2 \cdot 50) \cdot 37 = 100 \cdot 37 = 3700\).

Answer

a) \(1900\) b) \(48{,}000\) c) \(600\) d) \(3700\)

5192697

A product has two factors. Describe how the product changes in each case. a) Multiply the first factor by \(4\); leave the second factor unchanged. b) Multiply the first factor by \(10\) and divide the second factor by \(2\). c) Multiply both factors by \(3\).

Hints

- Find the scale factor caused by each change. - Multiply the scale factors when both factors change. - Test with simple factors if needed.

Solution

1. Multiplying one factor by \(4\) multiplies the product by \(4\). 2. The combined scale factor is \(10 \div 2 = 5\), so the product is multiplied by \(5\). 3. The combined scale factor is \(3 \cdot 3 = 9\), so the product is multiplied by \(9\).

Answer

a) The product is multiplied by \(4\). b) The product is multiplied by \(5\). c) The product is multiplied by \(9\).

5192717

Use \(12 \cdot 15 = 180\) to determine each product without calculating it from scratch. Briefly explain each change. a) \(24 \cdot 15\) b) \(12 \cdot 45\) c) \(6 \cdot 30\) d) \(4 \cdot 15\)

Hints

- Compare each factor with the corresponding factor in \(12 \cdot 15\). - Identify whether each factor was multiplied or divided by a scale factor. - Combine the effects when both factors change.

Solution

1. The first factor is doubled, so the product doubles: \(180 \cdot 2 = 360\). 2. The second factor is multiplied by \(3\), so the product is multiplied by \(3\): \(180 \cdot 3 = 540\). 3. The first factor is halved and the second is doubled. The scale factors cancel, so the product remains \(180\). 4. The first factor is divided by \(3\), so the product is divided by \(3\): \(180 \div 3 = 60\).

Answer

a) \(360\) b) \(540\) c) \(180\) d) \(60\)

5192797

Consider all whole-number division equations with positive integers and a quotient of exactly \(10\). How many such equations are there? Explain.

Hints

- Generate several examples with different positive divisors. - For a chosen divisor, determine the dividend that gives quotient \(10\). - Decide whether there is a greatest positive integer.

Solution

1. Choose any positive integer \(n\) as the divisor. 2. Use \(10n\) as the dividend. Then \(10n \div n = 10\). 3. Because there are infinitely many positive integers \(n\), there are infinitely many such division equations.

Answer

There are infinitely many. For every positive integer \(n\), \(10n \div n = 10\).

5195987

Rewrite the expression to make it easy to evaluate. Keep each operation sign with its number when reordering the terms. \(45 - 17 + 55 - 83 + 12\)

Hints

- Group terms to create multiples of \(100\). - Keep each subtraction sign with the number that follows it. - Find the total of the positive terms and the total being subtracted.

Solution

1. Group the positive terms and the amounts being subtracted: \((45 + 55 + 12) - (17 + 83)\). 2. Evaluate the groups: \(112 - 100 = 12\).

Answer

\(12\)

5196107

Use the commutative and associative properties to evaluate each product efficiently. a) \(4 \cdot 19 \cdot 5 \cdot 5\) b) \(8 \cdot 15 \cdot 125\) c) \(2 \cdot 13 \cdot 5 \cdot 10\) d) \(250 \cdot 7 \cdot 4\)

Hints

- Look for factors that make \(10\), \(100\), or \(1000\). - Reorder and regroup factors before multiplying. - Use powers of \(10\) to simplify the final multiplication.

Solution

1. For a), regroup: \((4 \cdot 5 \cdot 5) \cdot 19 = 100 \cdot 19 = 1900\). 2. For b), regroup: \((8 \cdot 125) \cdot 15 = 1000 \cdot 15 = 15{,}000\). 3. For c), regroup: \((2 \cdot 5) \cdot 10 \cdot 13 = 100 \cdot 13 = 1300\). 4. For d), regroup: \((250 \cdot 4) \cdot 7 = 1000 \cdot 7 = 7000\).

Answer

a) \(1900\) b) \(15{,}000\) c) \(1300\) d) \(7000\)

5196127

Use the distributive property to evaluate each expression. For each one, state whether you expanded an expression or factored out a common factor. a) \(8(100 + 4)\) b) \(17 \cdot 12 + 17 \cdot 8\) c) \((50 - 2) \cdot 6\) d) \(44 \cdot 15 - 34 \cdot 15\)

Hints

- When a factor is outside grouping symbols, distribute it to each term inside. - When two terms share a factor, factor it out. - Choose the form that makes the arithmetic easiest.

Solution

1. For a), expand: \(8(100 + 4) = 8 \cdot 100 + 8 \cdot 4 = 800 + 32 = 832\). 2. For b), factor out \(17\): \(17 \cdot 12 + 17 \cdot 8 = 17(12 + 8) = 17 \cdot 20 = 340\). 3. For c), expand: \((50 - 2) \cdot 6 = 50 \cdot 6 - 2 \cdot 6 = 300 - 12 = 288\). 4. For d), factor out \(15\): \(44 \cdot 15 - 34 \cdot 15 = (44 - 34)15 = 10 \cdot 15 = 150\).

Answer

a) \(832\); expanded b) \(340\); factored c) \(288\); expanded d) \(150\); factored

5196227

Evaluate each expression in two ways. Method 1: Evaluate inside the grouping symbols first, then divide. Method 2: Divide each term inside the grouping symbols, then combine the quotients. a) \((400 + 56) \div 8\) b) \((600 - 42) \div 6\) For each part, state which method you find easier and explain why.

Hints

- Check whether each term inside the grouping symbols is divisible by the divisor. - Compare one larger division with two simpler divisions. - Keep the addition or subtraction operation when splitting the quotient.

Solution

1. For a), Method 1 gives \(456 \div 8 = 57\). Method 2 gives \(400 \div 8 + 56 \div 8 = 50 + 7 = 57\). 2. For b), Method 1 gives \(558 \div 6 = 93\). Method 2 gives \(600 \div 6 - 42 \div 6 = 100 - 7 = 93\). 3. Method 2 may be easier because each term inside the grouping symbols is a convenient multiple of the divisor.

Answer

a) \(57\). Method 2 is easier because both \(400\) and \(56\) are divisible by \(8\). b) \(93\). Method 2 is easier because both \(600\) and \(42\) are divisible by \(6\).

5196357

Use the distributive property to rewrite and evaluate each expression efficiently. a) \(13 \cdot 57 + 13 \cdot 43\) b) \(99 \cdot 14\) c) \(62 \cdot 11\)

Hints

- In part a, identify the common factor. - Rewrite \(99\) using \(100\), and rewrite \(11\) using \(10\). - Use the form that creates the easiest products.

Solution

1. For a), factor out \(13\): \(13(57 + 43) = 13 \cdot 100 = 1300\). 2. For b), write \(99\) as \(100 - 1\): \((100 - 1)14 = 1400 - 14 = 1386\). 3. For c), write \(11\) as \(10 + 1\): \(62(10 + 1) = 620 + 62 = 682\).

Answer

a) \(1300\) b) \(1386\) c) \(682\)

5196367

Use the distributive property for division to rewrite and evaluate each expression efficiently. a) \(245 \div 5\) b) \((144 - 36) \div 12\) c) \(132 \div 11 + 88 \div 11\)

Hints

- Split a dividend into parts that are each divisible by the divisor. - When two quotients have the same divisor, consider combining their dividends. - Preserve the addition or subtraction operation when rewriting.

Solution

1. For a), split the dividend: \(245 \div 5 = 200 \div 5 + 45 \div 5 = 40 + 9 = 49\). 2. For b), divide each term: \(144 \div 12 - 36 \div 12 = 12 - 3 = 9\). 3. For c), combine the dividends: \((132 + 88) \div 11 = 220 \div 11 = 20\).

Answer

a) \(49\) b) \(9\) c) \(20\)

5196417

Write and evaluate an expression for this instruction: Multiply the product of \(125\) and \(9\) by the product of \(8\) and \(3\). Reorder and regroup the factors to calculate efficiently.

Hints

- Translate each phrase “the product of” into multiplication with grouping symbols. - Look for factors that multiply to \(1000\). - Use the commutative and associative properties to reorder and regroup.

Solution

1. The expression is \((125 \cdot 9)(8 \cdot 3)\). 2. Reorder and regroup the factors: \((125 \cdot 8)(9 \cdot 3)\). 3. Evaluate: \(1000 \cdot 27 = 27{,}000\).

Answer

\((125 \cdot 9)(8 \cdot 3) = (125 \cdot 8)(9 \cdot 3) = 27{,}000\)

5196437

Write and evaluate an expression for this instruction: Subtract the quotient of \(224\) and \(8\) from the quotient of \(1024\) and \(8\). Combine the quotients before dividing.

Hints

- Pay attention to which quotient is being subtracted from which. - Notice that both quotients have the same divisor. - Subtract the dividends before dividing.

Solution

1. The expression is \(1024 \div 8 - 224 \div 8\). 2. Since both quotients have the same divisor, combine the dividends: \((1024 - 224) \div 8\). 3. Evaluate: \(800 \div 8 = 100\).

Answer

\(1024 \div 8 - 224 \div 8 = (1024 - 224) \div 8 = 100\)

5196567

Use the distributive property to rewrite and evaluate each product. Decompose one factor as a convenient sum or difference. a) \(7 \cdot 102\) b) \(9 \cdot 98\) c) \(15 \cdot 21\)

Hints

- Look for a factor close to a multiple of \(10\) or \(100\). - Rewrite that factor as a sum or difference. - Distribute the other factor to both terms.

Solution

1. For a), write \(102\) as \(100 + 2\): \(7(100 + 2) = 700 + 14 = 714\). 2. For b), write \(98\) as \(100 - 2\): \(9(100 - 2) = 900 - 18 = 882\). 3. For c), write \(21\) as \(20 + 1\): \(15(20 + 1) = 300 + 15 = 315\).

Answer

a) \(7(100 + 2) = 714\) b) \(9(100 - 2) = 882\) c) \(15(20 + 1) = 315\)

5196577

Use the distributive property in reverse to factor out the common factor and evaluate each expression. a) \(14 \cdot 8 + 6 \cdot 8\) b) \(37 \cdot 12 - 27 \cdot 12\) c) \(5 \cdot 43 + 5 \cdot 57\)

Hints

- Identify the factor shared by both terms. - Place the remaining factors inside grouping symbols. - Add or subtract inside the grouping symbols before multiplying.

Solution

1. For a), factor out \(8\): \((14 + 6)8 = 20 \cdot 8 = 160\). 2. For b), factor out \(12\): \((37 - 27)12 = 10 \cdot 12 = 120\). 3. For c), factor out \(5\): \(5(43 + 57) = 5 \cdot 100 = 500\).

Answer

a) \(8(14 + 6) = 160\) b) \(12(37 - 27) = 120\) c) \(5(43 + 57) = 500\)

5197067

Use the distributive property to determine the missing numbers. The shapes represent new unknowns in each part. a) \(36 \cdot 4 + 14 \cdot 4 = (36 + \square) \cdot \triangle\) b) \((\triangle - 15) \cdot 7 = 25 \cdot 7 - 15 \cdot 7\) c) \(80 \div 8 - 16 \div 8 = (80 - 16) \div \square\) d) \(12 \cdot \square + 12 \cdot 8 = 12 \cdot 20\)

Hints

- Compare each expanded expression with its factored form. - Identify the common factor or divisor. - In part d, determine which number added to \(8\) gives \(20\).

Solution

1. For a), factor out \(4\): \((36 + 14) \cdot 4\). Thus, \(\square = 14\) and \(\triangle = 4\). 2. For b), compare with \((25 - 15) \cdot 7\). Thus, \(\triangle = 25\). 3. For c), both quotients have divisor \(8\), so \(\square = 8\). 4. For d), the factors inside the implied sum must satisfy \(\square + 8 = 20\), so \(\square = 12\).

Answer

a) \(\square = 14\); \(\triangle = 4\) b) \(\triangle = 25\) c) \(\square = 8\) d) \(\square = 12\)

5197437

Marie describes a number trick: 1. Choose a number. 2. Multiply it by \(25\). 3. Multiply the result by \(8\). 4. Divide that result by \(100\). Marie says, “I can recover your starting number by taking half of the final result.” a) Test her claim using a starting number of \(6\). b) Explain why the trick always works.

Hints

- Carry out the steps for \(6\). - Combine the constant factors \(25\), \(8\), and \(100\). - Represent the starting number with a variable.

Solution

1. Starting with \(6\): \(6 \cdot 25 \cdot 8 \div 100 = 12\). Half of \(12\) is \(6\), so the claim works in this case. 2. For any starting number \(n\), the steps give \(n \cdot 25 \cdot 8 \div 100\). 3. Since \(25 \cdot 8 = 200\) and \(200 \div 100 = 2\), the final result is \(2n\). Taking half returns \(n\).

Answer

a) The final result is \(12\), and half of \(12\) is \(6\). b) The operations are equivalent to multiplying the starting number by \(2\), so halving the result recovers the starting number.

5199487

Use the distributive property to rewrite and evaluate each expression. Make the use of the property clear in your work. a) \(7 \cdot 104\) b) \(18 \cdot 9\) c) \(35 \cdot 16 + 35 \cdot 4\)

Hints

- Rewrite a factor as a convenient sum or difference. - In part c, identify the common factor. - Choose a rewrite that creates multiples of \(10\) or \(100\).

Solution

1. For a), write \(104\) as \(100 + 4\): \(7(100 + 4) = 700 + 28 = 728\). 2. For b), write \(9\) as \(10 - 1\): \(18(10 - 1) = 180 - 18 = 162\). 3. For c), factor out \(35\): \(35(16 + 4) = 35 \cdot 20 = 700\).

Answer

a) \(7(100 + 4) = 728\) b) \(18(10 - 1) = 162\) c) \(35(16 + 4) = 700\)

5204137

The product \(14 \cdot 5\) can be evaluated in different ways. 1) Use the distributive property by writing \(14\) as \(10 + 4\). 2) Use the associative property by writing \(14\) as \(7 \cdot 2\). 3) Compare the two methods. Which is easier for this product, and why?

Hints

- Write \(14\) as both a sum and a product. - Apply the named property in each method. - Compare the number and difficulty of the steps.

Solution

1. Distributive method: \((10 + 4) \cdot 5 = 10 \cdot 5 + 4 \cdot 5 = 50 + 20 = 70\). 2. Associative method: \((7 \cdot 2) \cdot 5 = 7 \cdot (2 \cdot 5) = 7 \cdot 10 = 70\). 3. Both methods give \(70\). A reasonable comparison is that the second method is shorter because it creates the factor \(10\).

Answer

1) \((10 + 4) \cdot 5 = 70\) 2) \(7 \cdot (2 \cdot 5) = 70\) 3) The associative method is easier because it creates \(2 \cdot 5 = 10\) and uses fewer steps.

5204147

Two students evaluate \(25 \cdot 12\) using different strategies. Strategy A: Write \(12\) as \(10 + 2\). Strategy B: Write \(12\) as \(4 \cdot 3\). a) Evaluate the product using both strategies. b) Name the property used in each strategy.

Hints

- Expand the sum in Strategy A. - Regroup the factors in Strategy B. - Match each change to the property that permits it.

Solution

1. Strategy A: \(25(10 + 2) = 250 + 50 = 300\). This uses the distributive property. 2. Strategy B: \(25(4 \cdot 3) = (25 \cdot 4) \cdot 3 = 100 \cdot 3 = 300\). This uses the associative property.

Answer

a) Strategy A: \(25(10 + 2) = 300\). Strategy B: \((25 \cdot 4) \cdot 3 = 300\). b) Strategy A uses the distributive property; Strategy B uses the associative property.

5211226

Find the value of \(\square\) that makes the equation true: \(125 \cdot \square \cdot 4 = 1500\).

Hints

- Reorder and regroup the factors to make the multiplication easier. - Look for two factors whose product is a multiple of \(10\), \(100\), or \(1000\). - Then solve the resulting one-step equation.

Solution

1. Use the commutative and associative properties to group \(125\) and \(4\): \(125 \cdot 4 = 500\). 2. The equation becomes \(500 \cdot \square = 1500\). 3. Divide: \(\square = 1500 \div 500 = 3\).

Answer

\(\square = 3\)

5213877

Sarah tries to use the distributive property to calculate \(18 \cdot 102\): \(18(100 + 2) = 1800 + 2 = 1802\) a) Explain Sarah’s error. b) Show the correct calculation.

Hints

- A factor outside grouping symbols must multiply every term inside. - Identify the missing product in Sarah’s expansion. - Add the two correct partial products.

Solution

1. Sarah multiplied \(18\) by \(100\) but did not multiply \(18\) by \(2\). 2. The correct expansion is \(18(100 + 2) = 18 \cdot 100 + 18 \cdot 2\). 3. Therefore, the product is \(1800 + 36 = 1836\).

Answer

a) Sarah failed to distribute \(18\) to the second term. b) \(18(100 + 2) = 1800 + 36 = 1836\)

5217707

Rewrite the expression to make it easy to evaluate. Keep each subtraction sign with the number that follows it. \(154 + 78 - 54 - 28\)

Hints

- Look for pairs that make easy differences. - Keep each operation sign with its term when regrouping. - Add the two resulting differences.

Solution

1. Regroup as \((154 - 54) + (78 - 28)\). 2. Evaluate: \(100 + 50 = 150\).

Answer

\((154 - 54) + (78 - 28) = 150\)

5222137

At a charity walk, Class A raises \(\$x\). Class B raises \(20\%\) more than Class A. Class C raises \(\$50\) less than Class B. 1. Write an expression for the amount raised by Class B. 2. Write an expression for the amount raised by Class C. 3. Write and simplify an expression for the total amount raised by all three classes.

Hints

- Write the percent increase as a decimal. - Express each class's amount separately before adding. - Combine like terms when simplifying the total. - Distinguish variable terms from fixed amounts.

Solution

1. A \(20\%\) increase gives \(x+0.20x=1.20x\). 2. Class C raises \(1.20x-50\). 3. Add the three amounts: \(x+1.20x+(1.20x-50)=3.40x-50\).

Answer

1. \(1.20x\), or \(x+0.20x\) 2. \(1.20x-50\) 3. \(3.40x-50\)

5223986

Compare each pair of expressions. For each expression, identify the operation performed last and use it to name the structure of the entire expression. a) \(a^2+b^2\) and \((a+b)^2\) b) \(xy-z\) and \(x(y-z)\)

Hints

- Determine how parentheses change the order of operations. - The operation performed last names the structure of the entire expression. - Compare the outermost operation in each pair.

Solution

1. In \(a^2+b^2\), the powers are evaluated before the addition. The last operation is addition, so the expression is a sum. In \((a+b)^2\), the sum inside parentheses is evaluated before it is squared. The last operation is exponentiation, so the expression is a power. 2. In \(xy-z\), multiplication is performed before subtraction. The last operation is subtraction, so the expression is a difference. In \(x(y-z)\), the difference inside parentheses is evaluated before multiplication. The last operation is multiplication, so the expression is a product.

Answer

a) \(a^2+b^2\) is a sum; \((a+b)^2\) is a power. b) \(xy-z\) is a difference; \(x(y-z)\) is a product.

5224607

Use the distributive property. a) Fill in the missing number: \(6(20 + 7) = 6 \cdot 20 + 6 \cdot \dots\) b) Rewrite and evaluate \(8 \cdot 105\) efficiently.

Hints

- Match each term inside the grouping symbols with its product outside. - Decompose \(105\) using \(100\). - Distribute the factor to both terms.

Solution

1. In a), each addend inside the grouping symbols is multiplied by \(6\), so the missing number is \(7\). 2. For b), write \(105\) as \(100 + 5\): \(8(100 + 5) = 8 \cdot 100 + 8 \cdot 5 = 800 + 40 = 840\).

Answer

a) \(7\) b) \(8(100 + 5) = 800 + 40 = 840\)

5225349

A two-digit number has tens digit \(x\) and ones digit \(y\), where \(x\in\{1,\ldots,9\}\) and \(y\in\{0,\ldots,9\}\). 1. Write an expression for the value of the number. 2. Reverse the digits. Write an expression for the value of the reversed number. 3. Subtract the reversed number from the original number and simplify completely. 4. Use your result from part \(3\) to explain why the difference is always divisible by \(9\).

Hints

- A two-digit number can be written as the tens digit times \(10\), plus the ones digit. - Use parentheses around the entire reversed-number expression when you subtract. - What factored form would make divisibility by \(9\) clear?

Solution

1. The original number is \(10x+y\). 2. The reversed number is \(10y+x\). 3. The difference is \((10x+y)-(10y+x)=10x+y-10y-x=9x-9y=9(x-y)\). 4. Since \(x-y\) is an integer, \(9(x-y)\) is a multiple of \(9\). Therefore, the difference is divisible by \(9\).

Answer

1. \(10x+y\) 2. \(10y+x\) 3. \(9(x-y)\) 4. The difference is \(9\) times an integer, so it is divisible by \(9\).

5225507

A square garden bed is an \(s \times s\) array of unit squares, where \(s\) is a positive whole number. A border one unit square wide is placed around the outside of the bed. Two students write different expressions for the number of border squares: Expression 1: \(4s+4\) Expression 2: \(4(s+1)\) a) Use both expressions to find the number of border squares when \(s=5\). b) Use algebra to determine whether the expressions are equivalent. c) Explain how the arrangement leads to \(4s+4\). What does each part of the expression represent?

Hints

- Substitute \(s=5\) into each expression. - Apply the distributive property to Expression 2. - Count the noncorner squares along the four sides, then count the corners.

Solution

1. Using Expression 1, \(4\cdot5+4=24\). 2. Using Expression 2, \(4(5+1)=24\). 3. Apply the distributive property: \(4(s+1)=4s+4\). Therefore, the expressions are equivalent. 4. The term \(4s\) counts \(s\) border squares along each of the four sides, excluding the corners. The additional \(4\) counts the four corner squares.

Answer

a) Both expressions give \(24\) border squares. b) Yes. \(4(s+1)=4s+4\). c) \(4s\) counts the noncorner border squares along the four sides, and \(+4\) counts the four corners.

5225547

Two motorboats, A and B, are \(d\) miles apart on a river and travel directly toward each other. The current moves at \(v_s\) miles per hour. Boat A travels with the current and has a still-water speed of \(v_A\) miles per hour. Boat B travels against the current and has a still-water speed of \(v_B\) miles per hour, where \(v_B>v_s\). a) Write an expression for each boat's speed relative to the shore. b) Write and simplify an expression for their closing speed \(v_c\). c) How does the current speed affect the time until the boats meet? Justify your answer using part b).

Hints

- First find each boat's speed relative to the shore. - Speeds add when objects move directly toward each other. - Remove parentheses and combine like terms. - Identify which variables remain after simplification.

Solution

1. Boat A's shore speed is \(v_A+v_s\). Boat B's shore speed is \(v_B-v_s\). 2. Because they move toward each other, add their shore speeds: \(v_c=(v_A+v_s)+(v_B-v_s)\). 3. Simplify: \(v_c=v_A+v_s+v_B-v_s=v_A+v_B\). 4. The current cancels from the closing-speed expression. Therefore, under this model, the meeting time is \(t=\frac{d}{v_A+v_B}\) and does not depend on \(v_s\).

Answer

a) Boat A: \(v_A+v_s\); Boat B: \(v_B-v_s\) b) \(v_c=v_A+v_B\) c) The current speed does not affect the meeting time because its effects cancel in the closing speed.

5229199

Consider products of even and odd integers. a) Use algebraic expressions to show that the product of any even integer and any odd integer is always even. b) Use algebraic expressions to determine whether the product of two odd integers is even or odd. Justify your conclusion.

Hints

- Represent an even integer as \(2n\). - Represent an odd integer as \(2m+1\). - When expanding two binomials, multiply every term in one binomial by every term in the other. - Rewrite the final expression in the form \(2k\) or \(2k+1\).

Solution

1. Write an even integer as \(2n\) and an odd integer as \(2m+1\), where \(m\) and \(n\) are integers. 2. For part a), \(2n(2m+1)=4nm+2n=2(2nm+n)\). This is \(2\) times an integer, so the product is even. 3. For part b), \((2n+1)(2m+1)=4nm+2n+2m+1=2(2nm+n+m)+1\). This has the form \(2k+1\), so the product is odd.

Answer

a) The product is even because \(2n(2m+1)=2(2nm+n)\). b) The product is odd because \((2n+1)(2m+1)=2(2nm+n+m)+1\).

5229259

Choose any two-digit number. Subtract the sum of its digits from the number. Use variables to explain why the result is always divisible by \(9\).

Hints

- How can you write a two-digit number using its tens and ones digits? - What does “the sum of its digits” mean? - Write one expression for the entire subtraction. - Simplify until the divisibility is visible.

Solution

1. Let \(a\) be the tens digit and \(b\) the ones digit, where \(a\in\{1,\ldots,9\}\) and \(b\in\{0,\ldots,9\}\). The number is \(10a+b\). 2. The sum of its digits is \(a+b\). 3. Subtracting gives \((10a+b)-(a+b)=10a+b-a-b=9a\). 4. Since \(9a\) is a multiple of \(9\), the result is always divisible by \(9\).

Answer

The number is \(10a+b\), and its digit sum is \(a+b\). Their difference is \((10a+b)-(a+b)=9a\), which is divisible by \(9\).

5229269

Consider a three-digit number whose hundreds digit and ones digit are both nonzero. Reverse the order of its digits to form another three-digit number. Use variables and algebraic simplification to show that the difference between the original number and the reversed number is always a multiple of \(99\).

Hints

- Write a three-digit number as hundreds plus tens plus ones. - What happens to the middle digit when the digits are reversed? - Use parentheses when subtracting the reversed number. - Factor the simplified expression.

Solution

1. Let the digits be \(a\), \(b\), and \(c\), with \(a\ne0\) and \(c\ne0\). The original number is \(100a+10b+c\). 2. The reversed number is \(100c+10b+a\). 3. Their difference is \((100a+10b+c)-(100c+10b+a)\). 4. Simplifying gives \(100a+10b+c-100c-10b-a=99a-99c=99(a-c)\). 5. Since \(a-c\) is an integer, the difference is a multiple of \(99\).

Answer

The difference is \((100a+10b+c)-(100c+10b+a)=99(a-c)\). Therefore, it is always a multiple of \(99\).

5230047

A rectangle has positive side lengths \(a\) and \(b\), with \(0<x<b\). 1. Write an expression for the original perimeter \(P_{\text{old}}\). 2. The side of length \(a\) is increased by \(x\), while the side of length \(b\) is decreased by \(x\). Write and fully simplify the new perimeter \(P_{\text{new}}\). 3. Compare the two perimeter expressions. 4. Check both perimeters for \(a=40\,\text{m}\), \(b=25\,\text{m}\), and \(x=5\,\text{m}\).

Hints

- Start with the perimeter formula for a rectangle. - Replace the original side lengths with \(a+x\) and \(b-x\). - Simplify the expression inside the parentheses first. - Observe what happens to the \(x\)-terms.

Solution

1. The original perimeter is \(P_{\text{old}}=2(a+b)\). 2. The new side lengths are \(a+x\) and \(b-x\). Thus, \(P_{\text{new}}=2[(a+x)+(b-x)]\). 3. Inside the brackets, \(a+x+b-x=a+b\), so \(P_{\text{new}}=2(a+b)=P_{\text{old}}\). The perimeter is unchanged. 4. \(P_{\text{old}}=2\cdot(40+25)=130\,\text{m}\). The new sides are \(45\,\text{m}\) and \(20\,\text{m}\), so \(P_{\text{new}}=2\cdot(45+20)=130\,\text{m}\).

Answer

1. \(P_{\text{old}}=2(a+b)\) 2. \(P_{\text{new}}=2(a+b)\) 3. The perimeters are equal. 4. Both perimeters are \(130\,\text{m}\).

5230067

Six consecutive whole numbers begin with \(n+5\), where \(n\) is a whole number. a) Write expressions for all six numbers. b) Find and simplify the sum of the three smallest numbers and the sum of the three largest numbers. c) Subtract the sum of the three smallest numbers from the sum of the three largest numbers. What do you notice about the result?

Hints

- List all six expressions before forming the two sums. - Use parentheses around each entire sum when subtracting. - After simplifying, check whether the variable remains.

Solution

1. The six numbers are \(n+5\), \(n+6\), \(n+7\), \(n+8\), \(n+9\), and \(n+10\). 2. The sum of the three smallest is \((n+5)+(n+6)+(n+7)=3n+18\). 3. The sum of the three largest is \((n+8)+(n+9)+(n+10)=3n+27\). 4. Their difference is \((3n+27)-(3n+18)=9\). 5. The result is constant; it does not depend on \(n\).

Answer

a) \(n+5\), \(n+6\), \(n+7\), \(n+8\), \(n+9\), and \(n+10\) b) Smallest three: \(3n+18\); largest three: \(3n+27\) c) \(9\). The difference is the same for every value of \(n\).

5232629

Every odd integer can be written as \(2n+1\), where \(n\) is an integer. Show that the sum of two consecutive odd integers is always divisible by \(4\). 1. Write an expression for the odd integer immediately after \(2n+1\). 2. Add the two expressions. 3. Simplify and explain why the result is divisible by \(4\).

Hints

- Consecutive odd integers are \(2\) units apart. - Add the two algebraic expressions and combine like terms. - Factor the simplified expression to reveal a factor of \(4\). - A number is divisible by \(4\) when it can be written as \(4\) times an integer.

Solution

1. Consecutive odd integers differ by \(2\), so the next odd integer is \(2n+3\). 2. Their sum is \((2n+1)+(2n+3)\). 3. Simplify and factor: \((2n+1)+(2n+3)=4n+4=4(n+1)\). Because \(n+1\) is an integer, the sum is a multiple of \(4\).

Answer

The consecutive odd integers are \(2n+1\) and \(2n+3\). Their sum is \(4n+4=4(n+1)\), so it is always divisible by \(4\).

5235829

Explore the difference of the squares of two consecutive integers. a) Calculate the larger square minus the smaller square for the pairs \((1,2)\), \((4,5)\), and \((10,11)\). What pattern do you notice? b) Let \(n\) be the smaller integer. Write an expression for the difference between the square of \(n+1\) and the square of \(n\). c) Expand and simplify the expression. Explain why the result is always odd.

Hints

- “Difference of the squares” means square each number, then subtract. - Consecutive integers differ by \(1\). - Expand \((n+1)(n+1)\) using the distributive property. - An odd integer has the form \(2k+1\).

Solution

1. The examples are \(2^2-1^2=3\), \(5^2-4^2=9\), and \(11^2-10^2=21\). Each result is odd. 2. The general expression is \((n+1)^2-n^2\). 3. Expand: \((n+1)^2-n^2=n^2+2n+1-n^2=2n+1\). 4. Since \(2n\) is even for every integer \(n\), \(2n+1\) is always odd.

Answer

a) \(3\), \(9\), and \(21\); all are odd. b) \((n+1)^2-n^2\) c) The expression simplifies to \(2n+1\), which is always odd.

5238017

A savings balance \(K\) earns an annual interest rate of \(p\%\) for one year. Write an expression for the balance after one year. Then evaluate the expression for: 1) \(K = \$2500\), \(p = 2\) 2) \(K = \$12{,}000\), \(p = 1.4\)

Hints

- Find the interest as a percent of the original balance. - The ending balance must be greater than the beginning balance. - A growth factor can combine the original balance and the increase. - Substitute each set of values into your expression.

Solution

1. The balance after one year is \(K\left(1 + \frac{p}{100}\right)\). An equivalent expression is \(K + \frac{Kp}{100}\). 2. For case 1, \(\$2500\left(1 + \frac{2}{100}\right) = \$2500 \cdot 1.02 = \$2550\). 3. For case 2, \(\$12{,}000\left(1 + \frac{1.4}{100}\right) = \$12{,}000 \cdot 1.014 = \$12{,}168\).

Answer

Expression: \(K\left(1 + \frac{p}{100}\right)\) 1) \(\$2550\) 2) \(\$12{,}168\)

5238027

A store reduces an item’s original price \(x\) by \(r\%\). Write an expression for the sale price. Then evaluate it for: 1) \(x = \$45\), \(r = 20\) 2) \(x = \$129\), \(r = 15\)

Hints

- A discount subtracts part of the original price. - After a discount of \(r\%\), what percent of the original price remains? - Follow the order of operations when substituting values.

Solution

1. A discount of \(r\%\) leaves \(1 - \frac{r}{100}\) of the original price, so the sale price is \(x\left(1 - \frac{r}{100}\right)\). An equivalent expression is \(x - \frac{xr}{100}\). 2. For case 1, \(\$45\left(1 - \frac{20}{100}\right) = \$45 \cdot 0.80 = \$36\). 3. For case 2, \(\$129\left(1 - \frac{15}{100}\right) = \$129 \cdot 0.85 = \$109.65\).

Answer

Expression: \(x\left(1 - \frac{r}{100}\right)\) 1) \(\$36\) 2) \(\$109.65\)

5244659

A student claims that the sum of any three odd integers is always odd. Verify the claim by writing and simplifying an expression for the sum of three arbitrary odd integers.

Hints

- Represent each odd integer in the form \(2k+1\). - Add the three expressions and combine like terms. - Rewrite the result in the form \(2k+1\). - Every even number is divisible by \(2\).

Solution

1. Write the three odd integers as \(2a+1\), \(2b+1\), and \(2c+1\), where \(a\), \(b\), and \(c\) are integers. 2. Add and combine like terms: \((2a+1)+(2b+1)+(2c+1)=2a+2b+2c+3\). 3. Rewrite \(3\) as \(2+1\) and factor: \(2a+2b+2c+3=2(a+b+c+1)+1\). 4. Since \(a+b+c+1\) is an integer, the sum has the form \(2k+1\), so it is odd.

Answer

The claim is true. The sum can be written as \(2(a+b+c+1)+1\), which has the form of an odd integer.

5279497

A motorboat travels on a river. Its speed in still water is \(v\), and the river current is \(c\). 1) Write an expression for the difference between the boat's downstream speed and upstream speed. 2) Simplify the expression. What does the result show? 3) Check the result when \(v=22\) miles per hour and \(c=4\) miles per hour.

Hints

- Express the downstream and upstream speeds first. - Subtract the entire upstream expression. - Distribute the negative sign carefully. - Check whether one variable cancels.

Solution

1. The downstream speed is \(v+c\), and the upstream speed is \(v-c\). Their difference is \((v+c)-(v-c)\). 2. Simplifying gives \(v+c-v+c=2c\). The difference is twice the current speed and does not depend on the boat's still-water speed. 3. The downstream speed is \(22+4=26\) miles per hour, and the upstream speed is \(22-4=18\) miles per hour. Their difference is \(26-18=8\) miles per hour, which equals \(2\cdot 4=8\) miles per hour.

Answer

1) \((v+c)-(v-c)\) 2) \(2c\); the difference is twice the current speed. 3) \(8\) miles per hour

5279507

An airplane flies at a constant speed \(v\) in still air. A wind with speed \(w\) changes the airplane's ground speed. a) Write expressions for the ground speed with a tailwind, \(v_T\), and with a headwind, \(v_H\). b) Add \(v_T\) and \(v_H\), then divide by \(2\). Simplify and interpret the result. c) An airplane's ground speed is \(520\) miles per hour with a tailwind and \(440\) miles per hour with a headwind. Find the airplane's still-air speed and the wind speed.

Hints

- Add the wind speed for a tailwind and subtract it for a headwind. - Look for terms that cancel when the two expressions are added. - Simplify the sum before dividing by \(2\). - Compare either ground speed with the still-air speed to find \(w\).

Solution

1. With a tailwind, \(v_T=v+w\). With a headwind, \(v_H=v-w\). 2. Their mean is \(\frac{(v+w)+(v-w)}{2}=\frac{2v}{2}=v\). It equals the airplane's still-air speed. 3. The still-air speed is \(\frac{520+440}{2}=480\) miles per hour. 4. The wind speed is \(520-480=40\) miles per hour.

Answer

a) \(v_T=v+w\); \(v_H=v-w\) b) \(v\), the airplane's still-air speed c) Still-air speed: \(480\) miles per hour; wind speed: \(40\) miles per hour

5280849

Consider any three-digit number with hundreds digit \(h\), tens digit \(t\), and ones digit \(o\). a) Write an expression for the value of the number and an expression for the sum of its digits. b) Subtract the digit sum from the number and simplify. c) What is the greatest one-digit positive integer that always divides the result? Justify your answer using the simplified expression.

Hints

- Write the number as hundreds plus tens plus ones. - When subtracting the digit sum, distribute the negative sign to every term. - Look for the greatest common factor of the coefficients in the simplified expression.

Solution

1. The number is \(100h+10t+o\), and its digit sum is \(h+t+o\). 2. The difference is \((100h+10t+o)-(h+t+o)\). 3. Simplifying gives \(99h+9t\). 4. Factoring gives \(99h+9t=9(11h+t)\). Therefore, the result is always divisible by \(9\). 5. Since \(9\) is the greatest one-digit positive integer, it is the requested divisor.

Answer

a) Number: \(100h+10t+o\); digit sum: \(h+t+o\) b) \(99h+9t\) c) \(9\), because \(99h+9t=9(11h+t)\).

5317137

Use calculation trees 1 and 2. a) Write a numerical expression for each tree. Use grouping symbols so each expression preserves the tree structure. b) Evaluate both expressions by working through the trees step by step. Give the final value of each tree.

Figure for problem 531713

Hints

- Follow the connections in each tree to identify which operation happens first. - In Tree 1, evaluate the exponent before adding and multiplying. - Use grouping symbols when an operation must happen before another operation that would normally have higher precedence. - Review the sign rules for multiplying and dividing negative numbers. - Work from the upper branches toward the final box.

Solution

1. Tree 1 first evaluates \((-2)^3=-8\), then adds \(-8+7=-1\), and then multiplies by \(-4\). Its expression is \(((-2)^3+7)\cdot(-4)\), and its final value is \((-1)\cdot(-4)=4\). 2. Tree 2 first divides \(60\div(-5)=-12\), then adds that result to \(-100\). Its expression is \(-100+(60\div(-5))\), and its final value is \(-100+(-12)=-112\).

Answer

a) Tree 1: \(((-2)^3+7)\cdot(-4)\); Tree 2: \(-100+(60\div(-5))\) b) Tree 1: \(4\); Tree 2: \(-112\)

5352027

The calculation tree includes negative numbers. Write the numerical expression represented by the tree, then find its final value. Pay close attention to the sign rules.

Figure for problem 535202

Hints

- Evaluate the exponent before adding on the left branch. - Work through the tree from the branch operations toward the final operation. - Subtracting a negative number is the same as adding its opposite. - Use grouping symbols to preserve the operations shown in each branch.

Solution

1. The tree represents \((-20+6^2)+(10-(-15))\). 2. Evaluate the exponent and the left branch: \(6^2=36\), so \(-20+36=16\). 3. Evaluate the right branch: \(10-(-15)=10+15=25\). 4. Add the branch results: \(16+25=41\).

Answer

The expression is \((-20+6^2)+(10-(-15))\), and its value is \(41\).

5352387

Use the calculation tree shown. Find the missing values in the empty boxes, then write the corresponding numerical expression with grouping symbols.

Figure for problem 535238

Hints

- Evaluate the exponent before the addition. - Start with the box connected directly to the two values on the right branch. - Review the sign rule for division involving a negative number. - Use grouping symbols so the addition is performed before the division.

Solution

1. Evaluate the exponent: \(3^2=9\). 2. Evaluate the branch operation: \(9+3=12\). 3. Use that result in the final division: \(-60\div12=-5\). 4. Because the addition must happen before the division, the expression is \(-60\div(3^2+3)\).

Answer

The intermediate value is \(12\), the final value is \(-5\), and the expression is \(-60\div(3^2+3)\).

5352957

Write the numerical expression represented by the calculation tree and evaluate it using decimals.

Figure for problem 535295

Hints

- Evaluate the exponent before multiplying. - Start with the values at the ends of the upper branches and follow the tree toward the bottom. - Each operation combines the values directly connected to it. - Track the signs carefully when multiplying, adding, and dividing decimals.

Solution

1. Evaluate the exponent: \(2^2=4\). 2. Evaluate the multiplication branch: \(-0.4\cdot4=-1.6\). 3. Add that result to the other branch value: \(0.6+(-1.6)=-1.0\). 4. Divide by the final value: \(-1.0\div0.2=-5\). 5. The expression is \((0.6+(-0.4)\cdot2^2)\div0.2\).

Answer

The expression is \((0.6+(-0.4)\cdot2^2)\div0.2\), and its value is \(-5\).

5546957

A community workshop charges a one-time registration fee of \(\$18\). For each participant, the materials cost \(\$4\) and the meal costs \(\$6\). The total cost for \(n\) participants is \(18+4n+6n\). Rewrite the expression so the fixed cost and the total per-participant cost are immediately visible. Explain the meaning of each part of your new expression.

Hints

- Identify which terms depend on the number of participants. - Combine like terms that have the same variable. - Interpret the constant separately from the coefficient of \(n\).

Solution

1. Combine the two per-participant terms: \(4n+6n=10n\). 2. The equivalent expression is \(18+10n\). 3. The constant \(18\) is the one-time registration fee, and the coefficient \(10\) is the combined materials-and-meal cost for each participant.

Answer

\(18+10n\). The \(18\) is the fixed fee, and \(10\) is the cost per participant.

5546967

A greenhouse uses water at a constant rate of \(2.4\) gallons per hour. It runs the irrigation system for \(h\) hours on Monday and \(k\) hours on Tuesday. The total water used is \(2.4h+2.4k\). Rewrite the expression in a form that makes the total irrigation time visible, and explain what the factor \(2.4\) means in the new form.

Hints

- Look for the factor shared by both terms. - Ask what \(h+k\) represents in the situation. - Connect the shared numerical factor to the unit rate.

Solution

1. Factor the common rate: \(2.4h+2.4k=2.4(h+k)\). 2. The quantity \(h+k\) is the total irrigation time over the two days. 3. The factor \(2.4\) is the water-use rate in gallons per hour.

Answer

\(2.4(h+k)\). The factor \(2.4\) is the gallons-per-hour rate, and \(h+k\) is the total irrigation time.

5111877

Evaluate the expression. Use structure to simplify the numerator and denominator before doing the final division: \(\frac{12.5\cdot3.7+12.5\cdot4.3}{\left(\frac{2}{3}+\frac{1}{6}\right)\div\frac{1}{12}}\)

Hints

- Treat the numerator and denominator as separate expressions first. - Is there a common factor in both products in the numerator? - Simplify the denominator before doing the final division.

Solution

1. Factor the common factor in the numerator: \(12.5\cdot3.7+12.5\cdot4.3=12.5(3.7+4.3)=12.5\cdot8=100\). 2. Simplify the denominator: \(\frac{2}{3}+\frac{1}{6}=\frac{5}{6}\), so \(\frac{5}{6}\div\frac{1}{12}=\frac{5}{6}\cdot12=10\). 3. Divide: \(100\div10=10\).

Answer

\(10\)

5142297

Consider these two instructions. Expression A: Subtract the difference of \(15.2\) and \(6.4\) from their sum. Expression B: Find twice \(6.4\). a) Write an expression for each instruction and evaluate it. b) Explain why the results agree by simplifying Expression A for any two numbers \(x\) and \(y\).

Hints

- Evaluate each numerical expression carefully. - What happens to the signs inside parentheses when a minus sign is in front? - Use \(x\) for the first number and \(y\) for the second, then combine like terms.

Solution

1. Expression A is \((15.2 + 6.4) - (15.2 - 6.4)\). Its value is \(21.6 - 8.8 = 12.8\). 2. Expression B is \(2 \cdot 6.4 = 12.8\). 3. In general, \((x + y) - (x - y) = x + y - x + y = 2y\). 4. Therefore, subtracting the difference from the sum always gives twice the second number.

Answer

a) Expression A: \((15.2 + 6.4) - (15.2 - 6.4) = 12.8\); Expression B: \(2 \cdot 6.4 = 12.8\) b) \((x + y) - (x - y) = 2y\), so the result depends only on the second number and equals twice that number.

5178417

In \(m - s = d\), the minuend \(m\) is increased by \(10\). Determine how the subtrahend \(s\) must change so that the new difference has each property. a) The difference stays the same. b) The difference is \(15\) greater than before. c) The difference is \(5\) less than before.

Hints

- First determine the effect of increasing the minuend by \(10\). - Increasing the subtrahend lowers the difference; decreasing it raises the difference. - Work backward from the desired net change.

Solution

1. Increasing \(m\) by \(10\) first increases the difference by \(10\). To cancel that change, increase \(s\) by \(10\): \((m + 10) - (s + 10) = d\). 2. To make the total increase \(15\), the difference must increase by another \(5\). Decrease \(s\) by \(5\): \((m + 10) - (s - 5) = d + 15\). 3. To make the final difference \(5\) less than before, the change from the subtrahend must be \(-15\). Increase \(s\) by \(15\): \((m + 10) - (s + 15) = d - 5\).

Answer

a) Increase the subtrahend by \(10\). b) Decrease the subtrahend by \(5\). c) Increase the subtrahend by \(15\).

5186397

Consider the expression \(2500 - [750 - (150 - 50)]\). First evaluate the expression. Then determine how its value changes if each of the four numbers is increased by \(10\). Explain why.

Hints

- Evaluate the expression from the innermost grouping outward. - Track how each inner difference changes before considering the outer subtraction. - Recall what happens when both numbers in a subtraction increase by the same amount.

Solution

1. Evaluate from the innermost grouping outward: \(150 - 50 = 100\), \(750 - 100 = 650\), and \(2500 - 650 = 1850\). 2. Increasing both numbers in \(150 - 50\) by \(10\) leaves that difference unchanged. 3. The value of \(750 - (150 - 50)\) increases by \(10\), because only its first number has a net increase. 4. In the outer subtraction, both the first number and the value being subtracted increase by \(10\). Therefore, the final difference remains \(1850\).

Answer

The original value is \(1850\), and it remains unchanged.

5196247

Evaluate \((1100 - 55) \div 11\) in two ways: first by subtracting before dividing, and then by dividing each term before subtracting. Decide whether this statement is true or false: “When dividing a difference by a number, splitting the quotient is always more complicated.” Use your calculations to justify your answer.

Hints

- Calculate the expression both ways. - Compare the difficulty of the divisions in each method. - A single counterexample is enough to show that an “always” statement is false.

Solution

1. Subtract first: \((1100 - 55) \div 11 = 1045 \div 11 = 95\). 2. Divide each term first: \(1100 \div 11 - 55 \div 11 = 100 - 5 = 95\). 3. The statement is false. In this example, splitting the quotient is easier because both \(1100\) and \(55\) are convenient multiples of \(11\).

Answer

The value is \(95\). The statement is false because \(1100 \div 11 - 55 \div 11 = 100 - 5\) is simpler than \(1045 \div 11\).

5197447

Follow these steps: - Choose a whole number. - Multiply it by \(3\). - Add \(15\). - Divide the new sum by \(3\). - Subtract \(5\). Test the process with \(4\), \(10\), and \(25\). What pattern do you notice? Explain why it occurs.

Hints

- Carry out every step for each test number. - Represent the starting number with a variable. - Distribute the division by \(3\) across both terms in the numerator.

Solution

1. For \(4\): \((3 \cdot 4 + 15) \div 3 - 5 = 9 - 5 = 4\). 2. For \(10\): \((3 \cdot 10 + 15) \div 3 - 5 = 15 - 5 = 10\). 3. For \(25\): \((3 \cdot 25 + 15) \div 3 - 5 = 30 - 5 = 25\). 4. For any starting number \(n\), \((3n + 15) \div 3 - 5 = n + 5 - 5 = n\). The process always returns the starting number.

Answer

The final result always equals the starting number because \((3n + 15) \div 3 - 5 = n\).

5213867

Paul claims, “When dividing, you can always distribute over grouping symbols, whether the grouped expression is the dividend or the divisor.” He gives these examples: Example A: \((80 + 40) \div 10 = 80 \div 10 + 40 \div 10\) Example B: \(120 \div (4 + 2) = 120 \div 4 + 120 \div 2\) Evaluate both sides of each example. Decide whether Paul’s general claim is correct and explain the restriction.

Hints

- Evaluate the left and right sides separately. - Note whether the grouped expression is being divided or is doing the dividing. - One counterexample disproves an “always” claim.

Solution

1. In Example A, the left side is \(120 \div 10 = 12\), and the right side is \(8 + 4 = 12\). This equality is true. 2. In Example B, the left side is \(120 \div 6 = 20\), and the right side is \(30 + 60 = 90\). This equality is false. 3. Division can be distributed over a sum or difference in the dividend, but not over a sum or difference in the divisor.

Answer

Paul is incorrect. Example A is true, but Example B is false because \(20 \ne 90\). Division can distribute over a sum or difference in the dividend, not in the divisor.

5240477

A cleaning concentrate is \(k\%\) active ingredient by volume. To prepare a cleaning solution, \(1\,\text{L}\) of the concentrate is mixed with \(n\,\text{L}\) of water. Assume the volumes add. a) Write an expression \(c(n)\) for the percent of active ingredient in the finished mixture. b) Rewrite your expression in the form \(k\cdot r(n)\%\). Explain what the factor \(r(n)\) reveals about how much of the original concentration remains. c) If the amount of water is doubled from \(n\) liters to \(2n\) liters, write the new concentration and use the two denominator structures to explain why doubling the water does not generally cut the concentration in half. Confirm with \(k=20\) and \(n=1\).

Hints

- Separate the amount of active ingredient from the total mixture volume. - After finding the concentration, factor out \(k\) so the remaining multiplicative factor can be interpreted. - When water doubles, identify exactly which term in the total-volume denominator changes. - Compare the structure of \(1+2n\) with twice \(1+n\) before substituting numbers.

Solution

1. One liter of concentrate contains \(\frac{k}{100}\,\text{L}\) of active ingredient, and the total volume is \((1+n)\,\text{L}\). 2. Therefore, \(c(n)=\frac{k}{1+n}\%\). 3. Rewrite as \(c(n)=k\cdot\frac{1}{n+1}\%\). The factor \(\frac{1}{n+1}\) is the fraction of the original concentration that remains after dilution. 4. Doubling the water gives \(c(2n)=\frac{k}{1+2n}\%\). The denominator changes from \(1+n\) to \(1+2n\), not to \(2(1+n)\), because the original \(1\,\text{L}\) of concentrate is not doubled. 5. With \(k=20\) and \(n=1\), \(c(1)=\frac{20}{2}\%=10\%\), while \(c(2)=\frac{20}{3}\%\approx6.67\%\), not \(5\%\).

Answer

a) \(c(n)=\frac{k}{1+n}\%\) b) \(c(n)=k\cdot\frac{1}{n+1}\%\); the factor \(\frac{1}{n+1}\) shows the fraction of the original concentration that remains. c) \(c(2n)=\frac{k}{1+2n}\%\). Doubling water does not double the full denominator \(1+n\). For \(k=20\), \(n=1\): \(10\%\) becomes approximately \(6.67\%\), not \(5\%\).

5241487

Someone claims, “No matter what number you choose, if you add \(5\), double the result, and then subtract twice your original number, the answer is always \(10\).” a) Test the claim by writing and simplifying an expression. b) After the doubling step, how could you change the remaining directions so that the final result is always the original number \(x\), rather than \(10\)? Give one possible rule.

Hints

- Write one expression for the original rule. - Simplify to see why the result does not depend on \(x\). - After doubling, compare \(2x+10\) with the desired result \(x\). - What terms must be removed so that only \(x\) remains?

Solution

1. The expression for the claim is \(2(x+5)-2x\). 2. Distributing and combining like terms gives \(2x+10-2x=10\), so the claim is true. 3. After doubling, the expression is \(2x+10\). To leave only \(x\), subtract \(10\) and subtract one copy of \(x\). 4. One possible revised ending is: subtract \(10\), then subtract the original number. Algebraically, \(2x+10-10-x=x\).

Answer

a) \(2(x+5)-2x=10\), so the claim is true. b) One possible rule is: after doubling, subtract \(10\), then subtract the original number. This gives \(2x+10-10-x=x\).

5546977

A sponsor pays \(25\%\) of a hiking club's van-and-entry bill. Before the sponsorship, the bill is \(120+8n\) dollars for \(n\) hikers. Two equivalent expressions for the club's share are \(0.75(120+8n)\) and \(90+6n\). a) Explain what each form reveals about the situation. b) Show algebraically that the forms are equivalent. c) If \(n\) increases by \(5\), use the form that makes the change easiest to see and find the increase in the club's share.

Hints

- Interpret \(0.75\) as the fraction of the original bill the club still pays. - Distribute the multiplier to compare the two forms. - For the change in \(n\), look for the form that makes the per-hiker contribution visible.

Solution

1. The form \(0.75(120+8n)\) shows that the club pays \(75\%\) of the entire original bill after the sponsor pays \(25\%\). 2. Distribute \(0.75\): \(0.75\cdot120+0.75\cdot8n=90+6n\), so the forms are equivalent. 3. The form \(90+6n\) shows a discounted fixed cost of \(\$90\) and a discounted cost of \(\$6\) per hiker. 4. Five additional hikers increase the club's share by \(6\cdot5=\$30\).

Answer

a) \(0.75(120+8n)\) reveals the \(25\%\) sponsorship of the whole bill; \(90+6n\) reveals a \(\$90\) fixed share plus \(\$6\) per hiker. b) \(0.75(120+8n)=90+6n\) c) The club's share increases by \(\$30\).

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